All categories

4 years ago

How do I solve this? Can you show me step by step process.

Mathematics

1 answer:

tiny-mole [99]4 years ago

8 0

(Q14) To find the equation when given the gradient and one set of points (or coordinates), you have to solve for b. 
1. Identify the gradient and x- and y- intercepts.
m=3, x=1, y=5
2. Substitute them into the equation, y=mx+b.
5=3(1) + b
3. Solve for b.
5=3 + b
b=2
4. Rewrite the equation.
y=3x+ 2
(Q15) To find the equation when given two sets of points, you need to.
1. Find the gradient, using the formula, (x2-x1 / y2-y1)
(7,5) and (4, -7)
x1=7, x2= 4, y1=5, y2=-7
4-7/-7-5 = -3/-12 = 1/4
Therefore, gradient = 1/4
2. Substitute with one set of points (can be either (7,5) or (4,-7) in this example - question 15) into the equation, (like you did in question 14)
a) Identify the gradient and x- and y- intercepts.
m=1/4, x=7, y=5
b) Solve for b.
5=1/4(7) + b
5=7/4 + b
b=13/4 or 3 1/4
c) Rewrite the equation.
y=1/4x + 13/4

You might be interested in

Frankie factored the expression ls 14m + 21pm - 35mq as 7(2m + 3pm - 5mq) . She thinks she has factored out the greatest common

shtirl [24]

Answer:

Disagree, as m is common to all terms of the expression and thus, can be factored. The factored expression is 7m(2 + 3p - 5q)

Step-by-step explanation:

If a variable is common to all terms of an expression, it can be factored.

14m + 21pm - 35mq

Here, m is common to all terms of the expression, so the factored expression is:

$7m(\frac{14m}{7m} + \frac{21pm}{7m} - \frac{35mq}{7m}) = 7m(2 + 3p - 5q)$

Disagree, as m is common to all terms of the expression and thus, can be factored. The factored expression is 7m(2 + 3p - 5q)

7 0

3 years ago

100 points for right answer Identify the correct explanation for why the triangles are similar. Then find SQ and TP.

STALIN [3.7K]

Answer:

Second one

Step-by-step explanation:

They have one angle (R) in common

Also,

Angles T & S and P & Q are equal because they are corresponding angles.

RS/RT = SQ/TP

4/(4+4) = 2x/(x+6)

4x = x + 6

3x = 6

x = 2

SQ = 2(2) = 4

TP = 2 + 6 = 8

8 0

4 years ago

The area of a rectangle is 48x®y7

Jlenok [28]

Answer:

Step-by-step explanation:

Remember, area equals width x height so you can set up this equation.

$48x^8y^7=6x^2y^4 * w$

Solving for w,

$\frac{48x^8y^7}{6x^2y^4} =w$

Simplifying,

$8x^6y^3$

Thus, D is the answer

7 0

3 years ago

Use the form of the definition of the integral given in the theorem to evaluate the integral. ∫ 0 − 2 ( 7 x 2 + 7 x ) d x

Murrr4er [49]

Answer:

$\int _{-2}^07x^2+7xdx=\frac{14}{3}$

Step-by-step explanation:

The definite integral of a continuous function f over the interval [a,b] denoted by $\int\limits^b_a {f(x)} \, dx$ , is the limit of a Riemann sum as the number of subdivisions approaches infinity. That is,

$\int\limits^b_a {f(x)} \, dx=\lim_{n \to \infty} \sum_{i=1}^{n}\Delta x \cdot f(x_i)$

where $\Delta x = \frac{b-a}{n}$ and $x_i=a+\Delta x\cdot i$

To evaluate the integral

$\int\limits^{0}_{-2} {7x^{2}+7x } \, dx$

you must:

Find $\Delta x$

$\Delta x = \frac{b-a}{n}=\frac{0+2}{n}=\frac{2}{n}$

Find $x_i$

$x_i=a+\Delta x\cdot i\\x_i=-2+\frac{2i}{n}$

Therefore,

$\lim_{n \to \infty}\frac{2}{n} \sum_{i=1}^{n} f(-2+\frac{2i}{n})$

$\int\limits^{0}_{-2} {7x^{2}+7x } \, dx=\lim_{n \to \infty}\frac{2}{n} \sum_{i=1}^{n} 7(-2+\frac{2i}{n})^{2} +7(-2+\frac{2i}{n})$

$\lim_{n \to \infty}\frac{2}{n} \sum_{i=1}^{n} 7(-2+\frac{2i}{n})^{2} +7(-2+\frac{2i}{n})\\\\\lim_{n \to \infty}\frac{2}{n} \sum_{i=1}^{n} 7[(-2+\frac{2i}{n})^{2} +(-2+\frac{2i}{n})]\\\\\lim_{n \to \infty}\frac{14}{n} \sum_{i=1}^{n} (-2+\frac{2i}{n})^{2} +(-2+\frac{2i}{n})$ $\lim_{n \to \infty}\frac{14}{n} \sum_{i=1}^{n} (-2+\frac{2i}{n})^{2} +(-2+\frac{2i}{n})\\\\\lim_{n \to \infty}\frac{14}{n} \sum_{i=1}^{n} 4-\frac{8i}{n}+\frac{4i^2}{n^2} -2+\frac{2i}{n}\\\\\lim_{n \to \infty}\frac{14}{n} \sum_{i=1}^{n} \frac{4i^2}{n^2}-\frac{6i}{n}+2$

$\lim_{n \to \infty}\frac{14}{n} \sum_{i=1}^{n} \frac{4i^2}{n^2}-\frac{6i}{n}+2\\\\\lim_{n \to \infty}\frac{14}{n}[ \sum_{i=1}^{n} \frac{4i^2}{n^2}-\sum_{i=1}^{n}\frac{6i}{n}+\sum_{i=1}^{n}2]\\\\\lim_{n \to \infty}\frac{14}{n}[ \frac{4}{n^2}\sum_{i=1}^{n}i^2 -\frac{6}{n}\sum_{i=1}^{n}i+\sum_{i=1}^{n}2]$

We can use the facts that

$\sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}$

$\sum_{i=1}^{n}i=\frac{n(n+1)}{2}$

$\lim_{n \to \infty}\frac{14}{n}[ \frac{4}{n^2}\cdot \frac{n(n+1)(2n+1)}{6}-\frac{6}{n}\cdot \frac{n(n+1)}{2}+2n]\\\\\lim_{n \to \infty}\frac{14}{n}[-n+\frac{2\left(n+1\right)\left(2n+1\right)}{3n}-3]\\\\\lim_{n \to \infty}\frac{14\left(n^2-3n+2\right)}{3n^2}$

$\frac{14}{3}\cdot \lim _{n\to \infty \:}\left(\frac{n^2-3n+2}{n^2}\right)\\\\\mathrm{Divide\:by\:highest\:denominator\:power:}\:1-\frac{3}{n}+\frac{2}{n^2}\\\\\frac{14}{3}\cdot \lim _{n\to \infty \:}\left(1-\frac{3}{n}+\frac{2}{n^2}\right)\\\\\frac{14}{3}\left(\lim _{n\to \infty \:}\left(1\right)-\lim _{n\to \infty \:}\left(\frac{3}{n}\right)+\lim _{n\to \infty \:}\left(\frac{2}{n^2}\right)\right)\\\\\frac{14}{3}\left(1-0+0\right)\\\\\frac{14}{3}$

Thus,

$\int _{-2}^07x^2+7xdx=\frac{14}{3}$

5 0

4 years ago

Mr. Altamirano opens an account with a simple interest on an account with $1000 at 3% annually for 9 months. How much interest d

gladu [14]

Answer:

1022.50

Step-by-step explanation:

1000*.03= 30. (12mos interest)

30/4=7.50 quarterly interest

7.50x3 (9mos)= 22.50 interest

7 0

3 years ago